Inverse not always Unique for Non-Associative Operation
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Theorem
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\circ$ be a non-associative operation.
Then for any $x \in S$, it is possible for $x$ to have more than one inverse element.
Proof
Consider the algebraic structure $\struct {S, \circ}$ consisting of:
- The set $S = \set {a, b, e}$
- The binary operation $\circ$
whose Cayley table is given as follows:
- $\begin {array} {c|cccc} \circ & e & a & b \\ \hline e & e & a & b \\ a & a & e & e \\ b & b & e & e \\ \end {array}$
By inspection, we see that $e$ is the identity element of $\struct {S, \circ}$.
We also note that:
\(\ds \paren {a \circ a} \circ b\) | \(=\) | \(\ds e \circ b\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds b\) |
\(\ds a \circ \paren {a \circ b}\) | \(=\) | \(\ds a \circ e\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a\) |
and so $\circ$ is not associative.
Note further that:
\(\ds a \circ b\) | \(=\) | \(\ds e\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds b \circ a\) |
and also:
\(\ds a \circ a\) | \(=\) | \(\ds e\) |
So both $a$ and $b$ are inverses of $a$.
Hence the result.
$\blacksquare$
Also see
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 3.1$: Monoids: Exercise $(12)$